Question
The population of one type of fish p1 and another type of fish p2 are modeled by the equations P Subscript 1 Baseline equals 1000 left-parenthesis StartFraction 2 over 3 EndFraction right-parenthesis Superscript t Baseline and P Subscript 2 Baseline equals 1000 left-parenthesis StartFraction 1 over 3 EndFraction right-parenthesis Superscript t baseline where t is the time in years.
Which expression describes the relationship between the populations P Subscript 1 Baselineand P Subscript 2 Baseline in 3 years?
Which expression describes the relationship between the populations P Subscript 1 Baselineand P Subscript 2 Baseline in 3 years?
Answers
Answer
give equarion in p2 = fraction p1
Answered by
GPT 3.5
To find the relationship between the populations P1 and P2 in 3 years, we can substitute t = 3 into the equations:
P1 = 1000(2/3)^3 = 1000(8/27) = 296.296
P2 = 1000(1/3)^3 = 1000(1/27) = 37.037
Therefore, the relationship between the populations P1 and P2 in 3 years is:
P1 = 296.296 and P2 = 37.037
P1 = 1000(2/3)^3 = 1000(8/27) = 296.296
P2 = 1000(1/3)^3 = 1000(1/27) = 37.037
Therefore, the relationship between the populations P1 and P2 in 3 years is:
P1 = 296.296 and P2 = 37.037
Answered by
GPT 3.5
To find the relationship between the populations P1 and P2 in terms of a fraction, we can substitute the equations for P1 and P2 into the equation P2 = fraction * P1:
P1 = 1000(2/3)^3 = 1000(8/27) = 296.296
P2 = 1000(1/3)^3 = 1000(1/27) = 37.037
Therefore, the relationship between P1 and P2 in terms of a fraction is:
P2 = (37.037 / 296.296) * P1
P2 = (1/8) * P1
P2 = 1/8 * P1
P1 = 1000(2/3)^3 = 1000(8/27) = 296.296
P2 = 1000(1/3)^3 = 1000(1/27) = 37.037
Therefore, the relationship between P1 and P2 in terms of a fraction is:
P2 = (37.037 / 296.296) * P1
P2 = (1/8) * P1
P2 = 1/8 * P1
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